Question

Repeat Exercise $8$, but use the $ASA$ method.

Short answer

By $ASA$ method $\Delta OBT\cong \Delta EFG$

Step-by-step solution

Step 1. Given information.

A triangle is given.

Step 2. Construct the triangle ΔOBT.

A triangle congruent to $\Delta OBT$ using $ASA$ postulate,

Step 3. Concept used.

Consider the following ways-

Draw a line $L$.

Mark a point $F$on the line $L$.

Taking $F$as center draw an arc on the line $L$with radius of $BT$.

With $B$as center draw an arc which touches the sides $BO$ and $BT$.

With same radius draw an arc with $F$as center.

Then mark the distance between the points $P$ and $Q$.

Draw an arc of same measure with $S$ as center.

Draw a ray at the point of intersection, marked it as $J$.

With $T$as center, draw an arc which touches the sides $TB$ and $TO$.

With same radius draw an arc with $G$as center.

Then mark the distance between the points $U$ and $V$.

Draw an arc of same measure with $X$as center.

Draw a ray at the point of intersection, marked it as $K$.

Then mark the point of intersection as $E$.

Step 4. Apply ASA theorem.

Angle-Side-Angle theorem states that,

if two angles and any side of one triangle are congruent to two angles and any side of another triangle,

then the triangles are congruent.

Thus, $\Delta OBT$ is congruent to $\Delta EFG$.

Since, $\angle B\cong \angle F,\angle T\cong \angle G$ and $BT\cong FG$

then $\Delta OBT\cong \Delta EFG$.

Therefore, by $ASA$ method $\Delta OBT\cong \Delta EFG$.